Are closed rotationally symmetric surfaces which can be isometrically embedded into Euclidean space, always embedded as surfaces of revolution?

Question

Setup:
Consider a closed (compact without boundary) $2$-manifold $(\Sigma, h)$ with an isometric embedding into Euclidean $3$-space $\Phi \colon (\Sigma, h) \to (\mathbb{R}^3, \delta)$, that is $\delta|_{\Phi(\Sigma)}=h$.

Question:
If $(\Sigma, h)$ has $S^1$-symmetry, then does the image $\Phi(\Sigma)\subset \mathbb{R}^3$ automatically have $S^1$ symmetry as a subset of Euclidean space? I.e., does $\Phi(\Sigma)\subset \mathbb{R}^3$ look like a surface of revolution?

Phrased another way:
Let $iso(\Sigma, h)$ be the isometry group of the surface $(\Sigma, h)$ and let $iso(\mathbb{R}^3, \Phi(\Sigma), \delta)\subset iso(\mathbb{R}^3,\delta)$ be the set of isometries of Euclidean space $f\colon (\mathbb{R}^3, \delta) \to (\mathbb{R}^3, \delta)$ which preserve the surface as a subset, i.e. $f(\Phi(\Sigma)) =\Phi(\Sigma) \subset \mathbb{R}^3$. If $S^1\subset iso(\Sigma, h)$ is a subgroup, then is it true that $S^1 \subset iso(\mathbb{R}^3, \Phi(\Sigma), \delta)$ is also a subgroup?

Discussion:
I'm nearly certain that the answer to this question is "yes". Although my intuition for this problem mainly comes from the vague idea that embedded closed surfaces should be (infinitesimally) rigid and my inability to conceive of a potential counter example.

This problem is related to the Weyl Embedding Problem, the problem of isometrically embedding positive Gauss curvature $(S^2, h)$ into $(\mathbb{R}^3, \delta)$ so that the embedding to strictly mean convex. However the problem differs in three key ways: we assume that $(\Sigma, h)$ has $S^1$ symmetry, $(\Sigma, h)$ may have negative Gauss curvature at some points, and $\Sigma$ may be homeomorphic to either $S^2$ or $T^2$. My hope is the the symmetry is enough to compensate for the 'lack of rigidity' introduced by negative curvature.

Any surface $(\Sigma, h)$ with $S^1$ symmetry admits metric of the form $h = ds^2 + \rho^2(s) d\theta^2$. If there exists a function $z(s)$ such that $\dot{\rho}^2 + \dot{z}^2=1$, then (using cylindrical coordinates) $\Phi(s,\theta) := (\rho(s), z(s), \theta) \in \mathbb{R}_{\geq0} \times \mathbb{R}\times S^1 = \mathbb{R}^3$ describes an isometric embedding $\Phi\colon (\Sigma, h) \to (\mathbb{R}^3, \delta)$ which clearly satisfies the property that $S^1 \subset iso(\mathbb{R}^3, \Phi(\Sigma), \delta)$ as desired. The real question is two parts:

1. If $z(s)$ exists, then does $S^1 \subset iso(\mathbb{R}^3, \Psi(\Sigma), \delta)$ for every isometric embedding $\Psi\colon (\Sigma, h) \to (\mathbb{R}^3, \delta)$, not just the ones constructed through this method?
2. If $z(s)$ does not exists, then is it true that no such isometric embedding with the desired property exists?

Some things to keep in mind:

• There usually "boundary conditions" that need to be satisfied to guarentee $(\rho(s), z(s), \theta)$ is at the poles, but those are automatically satisfied because $(\Sigma, h)$ is assumed to be a closed smooth manifold.
• If the Guass curvature of $(\Sigma, h)$ were positive, then any embedding $\Phi \colon (\Sigma, h) \to (\mathbb{R}^3, \delta)$ into Euclidean space would be unique up to rigid motions, and the problem would be more or less trivial.
• If $(\Sigma, h)$ were allowed to be not closed then there would be trivial counter examples. Consider $(\Sigma, h) = (\mathbb{R}^2, \delta)$ embedded into $(\mathbb{R}^3, \delta)$ as a parabola cross a line (the taco embedding). Clearly $(\mathbb{R}^2, \delta)$ is rotationally symmetric but the taco isn't.
• Even in the positive curvature case embeddings are not unique for non-closed surfaces. A classic example is the fact that spherical caps (looks like a contact lens that goes in someone's eye) are not rigid. So somehow the closedness of $\Sigma$ is important to the proof.
• This question has a natural generalization which I state below. Perhaps seeing the problem in this generality will allow people to use techniques that I had not considered.

Generalized Statement:
Let $(\Sigma^{n-1}, h)$ for $n>2$ be a closed Riemannian manifold with an isometric embedding $\Phi \colon (\Sigma, h) \to (M^n, g)$. Does $\Phi$ induce an injective homomorphism of the isometry groups $\Phi_* \colon iso(\Sigma, h) \to iso(M, \Phi(\Sigma), g)$?
(Note that there are trivial counter examples when $n=2$ since all embeddings of $S^1$ are isometric.)

Answer 1

tl; dr: No.

For example, a cone of incident angle greater than $2\pi$ at the vertex embeds isometrically into Euclidean three-space, but not as a surface of rotation. (A cone of rotation has incident angle less than $2\pi$ at the vertex.)

If we take two congruent pieces of such a cone, embed them as parallel surfaces, smoothly join their boundary circles as shown, and smooth out their vertices in a circle-invariant way (not shown), we get a smooth circle-invariant metric on the sphere that isometrically embeds in Euclidean three-space but does not embed as a surface of revolution.

A comparable construction by "punching out neighborhoods of the vertices and smoothly joining the inner boundary circles" yields circle-invariant metrics on a torus that embed in Euclidean thre-space, but not as surfaces of rotation.

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In case it's of interest: If $M$ is a sphere, then (see for example A Symplectic Look at Surfaces of Revolution, l'Ens. Math. 49 (2003) 157–172):

1. Up to isometry, every smooth circle-invariant Riemannian metric of area $4\pi$ on $M$ corresponds to a smooth, positive real-valued function $\varphi$ on $[-1, 1]$ satisfying $|\varphi'(\pm1)| = 2$.
2. Such a metric embeds as a surface of revolution in Euclidean three-space if and only if $|\varphi'| \leq 2$ everywhere in $[-1, 1]$.

The linked paper gives qualitative (and quantitative) interpretation to the slope condition. Loosely, if $|\varphi'| > 2$, then "area accumulates faster than can be accomplished by a surface of revolution." In this framework, a "saddle cone" arises from a linear function of absolute slope greater than $2$.